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A vacuum cleaner has a measured sound level of 63.2 dB. What is the intensity of this sound?

User Itay Oded
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Answer: The intensity of the sound produced by the vacuum cleaner is 1.28 x 10^(-3) W/m^2.

Explanation: The intensity (I) of a sound wave is given by:

I = 10^(L/10) * I0

where L is the sound level in decibels (dB) and I0 is the reference intensity, which is 1 x 10^(-12) W/m^2.

Substituting the given values, we get:

I = 10^(63.2/10) * 1 x 10^(-12)

I = 10^(6.32) * 1 x 10^(-12)

I = 1.28 x 10^(-3) W/m^2

Therefore, the intensity of the sound produced by the vacuum cleaner is 1.28 x 10^(-3) W/m^2.

User Mehul Bhalala
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